## Thomas' Calculus 13th Edition

First Derivative: $y'=4x^{2}-1$ Second Derivative: $y''=8x$
Using the power rule First Derivative $y=\frac{4x^{3}}{3}-x$ $y'=\frac{(4)(3)x^{3-1}}{3}-(1)x^{1-1}$ $y'=4x^{2}-1$ Second Derivative $y'=4x^{2}-1$ $y''=(4)(2)x^{2-1}-(0)$ $y''=8x$ (the derivative of 1 will be 0 because of the rule of the constant function).